Considering the large semicircular arc $\Gamma$ in the upper half plane of $\oint{\frac{e^{iz}}{z}\,dz}$ using the Estimation Lemma with $z=Re^{i\theta}, \theta\in[0,\pi]$ means :
$$\sup\left|\frac{e^{iRe^{i\theta}}}{Re^{i\theta}}\right|= \sup\frac{|e^{iR(\cos\theta+i\sin\theta)}|}{R}=\sup\frac{|e^{-R\sin\theta}|}{R}$$
How can the numerator be simplified? In the range $\sin\theta$ can be positive, and as $R\to\infty$ the original semicircle wouldn’t vanish. If the integral inequality $\left|\int_{a}^{b}{f(x)\,dx}\right|\leq\int_{a}^{b}{|f(x)|\, dx}$ is used you can double the integral and use $\frac{\pi}{2}$ as the bound instead, and then use Jordan’s inequality. But the Estimation lemma follows from the integral inequality, so it must still be possible to bound the expression.
EDIT : Yes, Jordan's Lemma can be used, but I was more interested in how this extends to the definition of the Estimation Lemma and the absolute value (given it does not contain an integral, but rather an upper bound)
Essentially, you should use the Lebesgue convergence theorem (a.k.a. dominated convergence): If $\Gamma_R$ denotes the upper semicircle of Radius $R>0$, then (skipping everything you already did)
$$\left|\int\limits_{\Gamma_R} \frac{e^{iz}}{z}dz\right| \leq \cdots\leq \int\limits_0^\pi e^{-R\sin\theta}d\theta. \hspace{1cm}(*)$$
The family $(F_R)_{R>0}$ with $F_R : [0,\pi]\longrightarrow\mathbb{R}$, $\theta\longmapsto e^{-R\sin\theta}$ converges to $0$ pointwise almost everywhere on $[0,\pi]$ for $R\longrightarrow\infty$, and is dominated by the $L^1([0,\pi])$-function $\theta\longmapsto1$. Thus, the Lebesgue theorem can be applied and it follows that the integral on the right hand side of $(*)$ tends to $0$ for $R\longrightarrow 0$.
Alternatively (essentially doing the proof of the dominated convergence theorem by hand), for any $\varepsilon>0$, there exists a $R>0$ so that $|F_R(\theta)| \leq\frac{\varepsilon}{\pi}$ for all $\theta\in I_\varepsilon:=[0+\frac{\varepsilon}{2},\pi-\frac{\varepsilon}{2}]$. For all other values, i.e. for $\theta\in J_\varepsilon:=(I_\varepsilon)^c = [0,\frac{\varepsilon}{2})\cup(\pi-\frac{\varepsilon}{2},\pi]$, we still can use that $|F_R(\theta)|\leq 1$, so: $$\int\limits_0^\pi e^{-R\sin\theta}d\theta \leq \int\limits_{I_\varepsilon} \underbrace{F_R(\theta)}_{\leq\frac{\varepsilon}{\pi}}d\theta + \int\limits_{J_\varepsilon} \underbrace{F_R(\theta)}_{\leq 1} d\theta \leq \underbrace{\lambda(I_\varepsilon)}_{\leq \pi}\cdot\frac{\varepsilon}{\pi} + \underbrace{\lambda(J_\varepsilon)}_{\leq 2\frac{\varepsilon}{2}}\cdot1 \leq 2\varepsilon$$
Here, $\lambda$ denotes the $1$-dimensional Lebesgue measure, or in this case just the "length" of a real interval, i.e. $\lambda([a,b])=b-a$.